Question

Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the circle. In your final answer, state whether each point is on the interior, exterior, or circumference of the circle. Include your calculations as proof of each point’s location.
A. (-1, 1)

B. (10, 0)

C. (4, -8)

Answer 1

Hello :
 an equation of the circle Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²  
In this exercice : (x-3)²+y²=7²   ....a =3 and b=0 ( the center is :w(3,0))  ...    r=7
The distance between each point in each case is calculated and compared with the length of the radius
 1 ) A (-1, 1) :  Aw² = (-1-3)²+1² =17.....  Aw < r : interior the circle
  2 )  B(10, 0) :Bw² =(10-3)²+0² =49 ......r = Bw :circumference of the circle
 3 )  C(4, -8) :Cw² = (4-3)²+(-8)² =65..... Cw > r  : exterior  the circle

Answer 2

Answer:

The point (-1, 1) lies in interior of circle.

The point (10,0) lies on the circumference of circle.

The point (4,-8) lies on the exterior of circle.

Step-by-step explanation:

Given : The circle with the equation $(x - 3)^2+y^2 =49$

We have to  determine the location of each point with respect to the graph of the circle. also state whether each point is on the interior, exterior, or circumference of the circle.

Interior point of a circle :

A point is said to be interior when the distance of point from the center is less than the radius.

Exterior point of a circle :

A point is said to be exterior when the distance of point from the center is greater than the radius.

Circumference of a circle :

A point is said to be at circumference when the distance of point from the center is equal the radius.

For the point (-1, 1)

Put in the given equation of circle $(x - 3)^2+y^2 =49$

Consider LHS of equation

$(x - 3)^2+y^2$

put x = -1 and y = 1

$(-1-3)^2+1^2=16+1 = 17$

So, the point (-1, 1) lies in interior of circle.

For the point (10, 0)

Put in the given equation of circle $(x - 3)^2+y^2 =49$

Consider LHS of equation

$(x - 3)^2+y^2$

put x = 10 and y = 0

$(10-3)^2+0^2=49=49$

So, the point (10,0) lies on the circumference of circle.

For the point (4,-8)

Put in the given equation of circle $(x - 3)^2+y^2 =49$

Consider LHS of equation

$(x - 3)^2+y^2$

put x = 4 and y = -8

$(4-3)^2+(-8)^2=1+64=65>49$

So, the point (4,-8) lies on the exterior of circle.